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The conical pendulum is in steady circular motion with constant angular velocity omega as shown in Fig. Mass of the particle is M and string makes angle alpha with vertical. (a) Find angular momentum of the particle about the center C of the circle and the hinge O about which the string is attached. (b) Also check whether L is changing or is constant. |
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Answer» Solution :(a) Conceptualize/Classify: We start by finding `L_(C)`, angular momentum of mass M about point C. We can see from Fig. that velocity of the particle is always going to be perpendicular to the line joining particle with the center C. Also, both vectors r and v are lying on horizontal (xy) plane, so the angular momentum, `L_(C)` will be in the positive z direction. It has magnitude `|r_(BOT)||p|=|r||p|=rp`, where r is the radius of the circular motion as we can see from Fig. Calculation: The magnitude of momentum is given by `|p|=Mv=Mromega` The angular momentum about the center C of the circle `L_(C)=rxxp` = `Mr^(2)omegahatk` Now LET us evaluate the angular momentum `L_(O)` about the point O located at the hinge. From Fig. `vecL_(0)=vecrxxvecp` Velocity of the particle is perpendicular to the line OA as SHOWN in the figure. `|L_(O)|=|r.xxp||r.||p|=L|p|` or `|L_(O)|=Mlromega` where `|r.|=l`, is the length of the string. (b) Calculation: We must note that `L_(C)` is constant, both in magnitude and direction. The magnitude of `L_(O)` is constant, but its direction is not constant as can be seen from Fig. `L_(O)` is always perpendicular to the plane containing r. and p. As the particle moves, the plane containing r. and p vector hinges, thus changing the direction of `L_(O)` at different times. As the bob swings around, `L_(O)` rotates as shown in fig. The vertical component of `L_(O)` is not moving, but the horizontal component rotates in the circle with the bob. We can draw the angular momentum at any position we CHOOSE, only the magnitude and direction of L have to be same. 
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